Answer:
Molar mass of MgCl2 is 95 g/mol
Mg = 24 g/mol and Cl = 35.5 ×2 = 71 g/mol
moles = mass given/ molar mass
= 2.7/95 = 0.028 mol
volume = 250/1000 = 0.25 dm3 (ml is the same as dm3)
molarity of MgCl2 = moles/volume
= 0.028/0.25
= 0.112 mol/dm3
The forensic technician at a crime scene has just prepared a luminol stock solution by adding 17.0 \({\rm g}\) of luminol into a total volume of 75.0 \(\rm mL\) of \(\rm H_2O\). What is the molarity of the stock solution of luminol
Answer:
1.28 M
Explanation:
Step 1: Given data
Mass of luminol (solute): 17.0 g
Volume of water: 75.0 mL (this is also the volume of solution)
Step 2: Calculate the moles corresponding to 17.0 g of luminol
The molar mass of luminol is 177.16 g/mol.
17.0 g × 1 mol/177.16 g = 0.0960 mol
Step 3: Calculate the molarity of the solution
We will use the definition of molarity
M = moles of solute / liters of solution
M = 0.0960 mol / 0.0750 L = 1.28 M
If 50.0 g of sulfuric acid and 40.0 grams of barium chloride are mixed, how many grams of sulfuric acid and how many grams of barium chloride remain after the double replacement reaction is complete?
After the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.
First, we will write the balanced equation for the reaction
H₂SO₄ + BaCl₂ → BaSO₄ + 2HCl
This means 1 mole of BaCl₂ is needed to react completely with 1 mole of H₂SO₄ to give 1 mole of BaSO₄ and 2 moles of HCl
From the question, 50.0g of sulfuric acid is mixed with 40.0 grams of barium chloride. To determine the quantity of each substance remaining after the complete reaction, we will first determine the number of moles present in each of the reactant.
For H₂SO₄
mass = 50.0g
Molar mass = 98.079 g/mol
From the formula
Number of moles = Mass / Molar mass
∴ Number of moles of H₂SO₄ = 50.0g / 98.079 g/mol
Number of moles of H₂SO₄ = 0.5098 mol
For BaCl₂
mass = 40.0 g
Molar mass = 208.23 g/mol
∴ Number of moles of BaCl₂ = 40.0g / 208.23 g/mol
Number of moles of BaCl₂ = 0.1921 mol
Since the number of moles of H₂SO₄ is more than that of BaCl₂, then H₂SO₄ is the excess reagent and BaCl₂ is the limiting reagent (that is, it will be used up completely during the reaction)
From the equation, 1 mole of H₂SO₄ is needed to completely react with 1 mole of BaCl₂
∴ 0.1921 mol of H₂SO₄ will be needed to completely react with 0.1921 mol of BaCl₂.
Therefore, after the reaction is complete, 0 mole (i.e 0 grams) of BaCl₂ will remain and (0.5098 mole - 0.1921 mole) of H₂SO₄ will remain.
Number of moles H₂SO₄ that will remain = 0.5098 mole - 0.1921 mole = 0.3177 moles
Now, we will convert this to grams
From the formula
Mass = Number of moles × Molar mass
Mass of H₂SO₄ that will remain = 0.3177 moles × 98.079 g/mol
Mass of H₂SO₄ that will remain = 31.1597 g
Mass of H₂SO₄ that will remain ≅ 31.16 g
Hence, after the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.
Learn more here: https://brainly.com/question/6857557
A student was asked to determine the percent mass of sodium nitrate in a mixture of sodium nitrate (NaNO3) and calcium carbonate (CaCO3). The mass of the mixture used was 3.2 g. The student extracted NaNO3 from the mixture with water and separated the insoluble CaCO3 from the solution by filtration. After evaporating the filtrate, the student recovered and dried the NaNO3, and found that it weighted 0.45 g. The student dried the insoluble residue of CaCO3 and found that it weighted 2.23 g. Calculate the percent mass of NaNO3 in the mixture and round your final answer to the correct number of significant figures.
Answer:
自分の仕事をする translate to english
Explanation:
Kc is 1.67 × 1020 at 25°C for the formation of iron(III) oxalate complex ion: Fe3+(aq) + 3 C2O42-(aq) ⇌ [Fe(C2O4)3]3-(aq). If 0.0800 M Fe3+ is initially mixed with 1.00 M oxalate ion, what is the concentration of Fe3+ ion at equilibrium?
Answer:
hello? are you still here? reply if you are
A compound with a molecular weight of about 64.47 g/mol was found to be 18.63 % of C, 1.56 % of H, 24.82 % of O, and 54.99 % of Cl by mass. Determine the molecular formula and draw the Lewis structure showing an accurate 3-D perspective. *Show your calculations
Answer:
See detailed explanation.
Explanation:
Hey there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly calculating the moles of each element, assuming those percentages are masses, so that we divide by their molar masses:
[tex]C=\frac{18.63}{12.01}=1.55\\\\H=\frac{1.56}{1.01} =1.55\\\\O=\frac{24.82}{16.00}=1.55\\\\Cl=\frac{54.99}{35.45}=1.55[/tex]
Then, we divide all of them by 1.55 to realize the empirical formula is:
[tex]CHOCl[/tex]
Whose molar mass is 64.47 g/mol, and therefore, since the molar mass of these two is the same, we infer the molecular formula is also CHOCl.
The Lewis structure is shown on the attached document, whereas, the central atom is C and it does complete its octet as well as both O and Cl.
Regards!
If ionization energy of hydrogen atom is 13.6 eV then its ionization potential will be
Ionization potential and ionization energy are two terms used to describe the same thing.
The ionization potential of hydrogen atom is 13.6 eV
The ionization potential is the energy that is required to remove an electron from the neutral atom. It is the same as the ionization energy.
From the question, we can see that the ionization energy of the hydrogen atom is 13.6 eV, it also means that the ionization potential of the hydrogen atom is also 13.6 eV.
Therefore, If ionization energy of hydrogen atom is 13.6 eV then its ionization potential will be 13.6 eV
For more information see:
https://brainly.com/question/16243729
all question are compulsory
Answer:
is this question or you just asking I can't understand.
The Nernst equation at 20oC is:
Eion= 58 millvolts/z. [log10 (ion)out/(ion)in]
Calculate the equilibrium potential for Cl- if the concentration of Cl- outside of the cell is 100 and the concentration inside of the cell is 10 mmol/liter.
a. 58 millivolts
b. +58 millivolts
c. -116 millivolts
d. 0
Answer:
a. -58 millivolts
Explanation:
The given Nernst equation is:
[tex]E_{ion} = 58 millivolts /z \Big[ log_{10} \Big( \dfrac{[ion]_{out}}{[ion]_{in}}\Big) \Big]}[/tex]
The equilibrium potential given by the Nernst equation can be determined by using the formula:
[tex]E_{Cl^-} = \dfrac{2.303*R*T}{ZF} \times log \dfrac{[Cl^-]_{out}} {[Cl^-]_{in}}[/tex]
where:
gas constant(R) = 8.314 J/K/mol
Temperature (T) = (20+273)K
= 298K
Faraday constant F = 96485 C/mol
Number of electron on Cl = -1
[tex]E_{Cl^-} = \dfrac{2.303*8.314*298} {(-1)*(96845)} \times log \dfrac{100} {10}[/tex]
[tex]E_{Cl^-} = - 0.05814 \ volts[/tex]
[tex]\mathsf{E_{Cl^-} = - 0.05814 \times 1000 \ milli volts}[/tex]
[tex]\mathsf{E_{Cl^-} \simeq - 58\ milli volts}[/tex]
Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.
Answer:
[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.
Explanation:
Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.
Analysis:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷
C(i) 0.115M 0 0
ΔC -x +x +x
C(eq) 0.115M - x x x
≅ 0.115M
Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M
= 4.3 x 10⁻⁷ => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.
In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion concentration, the hydroxide ion concentration is then calculated from
[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.
________________________________________________________
NOTE: The 2.32 x 10⁻⁴M value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.
The concentration of hydroxide ion of given solution is 4.3 x 10⁻¹¹M.
How we calculate the [OH⁻]?We can calculate the concentration of hydroxide ions as follow:
[OH⁻][H⁺] = 10⁻¹⁴
Given chemical reaction with ICE table shown as below:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq)
Initial: 0.115 0 0
Change: -x +x +x
Equilibrium: 0.115-x +x +x
Given that, Ka = 4.3 x 10⁻⁷
Equilibrium constant for this reaction is written as:
Ka = [H⁺][HCO₃⁻]/[H₂CO₃]
4.3 x 10⁻⁷ = x² / 0.115
x = 2.32 x 10⁻⁴M = [H⁺]
Now we calculate the concentration of hydroxide ion as:
[OH⁻][H⁺] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴ / 2.32 x 10⁻⁴ = 4.3 x 10⁻¹¹M
Hence, value of [OH⁻] is 4.3 x 10⁻¹¹M.
To know more about pH & pOH, visit the below ink:
https://brainly.com/question/24595796
For a different reaction, the plot of the reciprocal of concentration versus time in seconds was linear with a slope of 0.056 M-1 s -1 . If the initial concentration was 2.2 M, calculate the concentration after 100 seconds. Show your work.
Answer:
[tex]C_t=0.165M[/tex]
Explanation:
From the question we are told that:
Slope [tex]K=0.056 M-1 s -1[/tex]
initial Concentration [tex]C_1=2.2M[/tex]
Time [tex]t=100[/tex]
Generally the equation for Raw law is mathematically given by
[tex]\frac{1}{C}_t=kt+\frac{1}{C}_0[/tex]
[tex]\frac{1}{C}_t=0.056*100+\frac{1}{2.2}_0[/tex]
[tex]C_t=0.165M[/tex]
The second-order reaction is the reaction that depends on the reactants of the first or the second-order reaction. The concentration after 100 seconds will be 0.165 M.
What is the specific rate constant?The specific rate constant (k) of the second-order reaction is given in L/mol/s or per M per s. It is the proportionality constant that gives the relation between the concentration and the rate of the reaction.
Given,
Slope (k)= 0.056 per M per s
Initial concentration of the reactant [tex](\rm C_{1})[/tex] = 2.2 M
Time (t) = 100 seconds
The concentration of the reaction after 100 seconds can be given by,
[tex]\rm \dfrac{1}{C_{t}} = kt + \dfrac{1}{C_{1}}[/tex]
Substitute values in the above equation:
[tex]\begin{aligned} \rm \dfrac{1}{C_{t}} &= 0.056 \times 100 + \dfrac{1}{2.02}\\\\&= 0.165 \;\rm M\end{aligned}[/tex]
Therefore, after 100 seconds the concentration is 0.165 M.
Learn more about the order of reaction here:
https://brainly.com/question/14810717
Which type of organic compound is shown below?
A. Carboxylic acid
B. Ester
C. Amine
D. Alcohol
Answer:
I think its A maybe am not sure
Carbon dioxide gas is collected at 27.0 oC in an evacuated flask with a measured volume of 30.0L. When all the gas has been collected, the pressure in the flask is measured to be 0.480atm. Calculate the mass and number of moles of carbon dioxide gas that were collected.
Answer:
[tex]M_{CO_2}= 25.7g[/tex]
Explanation:
From the question we are told that:
Temperature [tex]T=27.0[/tex]
Volume [tex]V=30L[/tex]
Pressure [tex]P=0.480atm[/tex]
Generally the equation for Ideal gas is mathematically given by
PV=nRT
Therefore
[tex]n=\frac{0.480 x 30}{0.08205 x 300}[/tex]
[tex]n=0.59moles[/tex]
Generally Mass of CO2 is given as
[tex]M_{CO_2}= 0.59 * 44 g/mol[/tex]
[tex]M_{CO_2}= 25.7g[/tex]
Kamal was told by his mother to pour water through a thin cloth into another container to further purify the water.
a) What do you think will happen to the mud and sand when pouring the water? b)Do you think the water filtered by Kamal is safe to drink?
a].When sand is added to water it either hangs in the water or forms a layer at the bottom of the container. Sand therefore does not dissolve in water and is insoluble. It is easy to separate sand and water by filtering the mixture.
b]. The water filtered by kamal is not safe to drink .
If you run out of water, or cannot carry enough water with you for your entire trip, you may need to source drinking water from natural water sources.
3. Suppose you wanted to design an experiment to test the composition of a mixture that includes sodium phenoxide (NaC6H5O). You know that this solid mixture contains both the NaC6H5O and some inert NaCl, but do not know how much of each is present. You decide to test the composition by titrating with 0.100-M HCl. a. If a 1.000-g sample is 25% NaC6H5O by mass, how many mL of 0.100-M HCl would be required to reach the equivalence point of the titration
Answer:
21.5mL of a 0.100M HCl are required
Explanation:
The sodium phenoxide reacts with HCl to produce phenol and NaCl in a 1:1 reaction.
To solve this question we need to find the moles of sodium phenoxide. These moles = Moles of HCl required to reach equivalence point and, with the concentration, we can find the needed volume as follows:
Mass NaC6H5O:
1.000g * 25% = 0.250g NaC6H5O
Moles NaC6H5O -116.09g/mol-
0.250g NaC6H5O * (1mol/116.09g) = 2.154x10⁻³ moles = Moles of HCl required
Volume 0.100M HCl:
2.154x10⁻³ moles HCl * (1L/0.100mol) = 0.0215L =
21.5mL of a 0.100M HCl are requiredPLEASE HELP ASAP
A total of 132.33g C3H8 is burned in 384.00 g O2. Use the following questions to determine the amounts of products formed.
• How many grams of CO2 and H2O will be produced? (2 points)
b. If the furnace is not properly adjusted, the products of combustion can include other gases, such as CO and unburned hydrocarbons. If only 269.34 g of CO2 were formed in the above reaction, what would the percent yield be? (2 points)
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 6.90kg of water at 34.7 degrees C . During the reaction 57.1kJ of heat flows out of the bath and into the flask.
Calculate the new temperature of the water bath. You can assume the specific heat capacity of water under these conditions is 4.18J.g^(-1).K^(-1) . Round your answer to significant digits.
Answer:
[tex]T_2= 36.7 \textdegree C[/tex]
Explanation:
Mass of Water [tex]m_w=6.90kg[/tex]
Temperature [tex]T=34.7 degrees[/tex]
Heat Flow [tex]H=57.1kJ[/tex]
Specific heat capacity of water [tex]\mu= 4.18J.g^(-1).K^(-1)[/tex]
Generally the equation for Final Temperature is mathematically given by
[tex]M*\mu *T_1 + Q = M*\mu *T_2[/tex]
[tex]T_2=\frac{M*\mu *T_1 + Q }{M*\mu}[/tex]
Therefore
[tex]T_2=\frac{6.90*4.18*34.7 + 57.1}{6.90*4.18}[/tex]
[tex]T_2= 36.7 \textdegree C[/tex]
Consider the following titration for these three questions:
1.00 L of 2.00 M HCl is titrated with 2.00 M NaOH.
a. How many moles of acid are equal to one equivalent in this titration?
b. How many moles of HCl are found in solution at the halfway point of the titration?
c. How many liters of base will be needed to reach the equivalence point of the titration?
Answer:
a. 1 mole of acid is equal to one equivalent.
b. 1.00 moles of HCl are found.
c. 1L of 2.00M NaOH is needed to reach the equivalence point
Explanation:
HCl reacts with NaOH as follows:
HCl + NaOH → NaCl + H2O
Where 1 mole of HCl reacts with 1 mole of NaOH. The reaction is 1:1
a. As the reaction is 1:1, 1 mole of acid is equal to one equivalent
b. The initial moles of HCl are:
1.00L * (2.00moles HCl / 1L) = 2.00 moles of HCl
At the halfway point, the moles of HCl are the half, that is:
1.00 moles of HCl are found
c. At equivalence point, we need to add the moles of NaOH needed for a complete reaction of the moles of HCl. As the moles of HCl are 2.00 and the reaction is 1:1, we need to add 2.00 moles of NaOH, that is:
2.00moles NaOH * (1L / 2.00mol) =
1L of 2.00M NaOH is needed to reach the equivalence point
Critique this statement: Electrons can exist in any position
outside of the nucleus.
Answer:
However, there has to be 2 electrons on the first shell, and 8 on the others.
Explanation:
Hope this helps :)
If atom X had 3 valence electrons and atom Y had 7 valence electrons, the correct chemical formula for the ionic compound they would form is Choices: A) XY3 B) X3Y6 C) X2Y3 D) X3Y7
Answer:
A) XY3
In both of the atom, they want to have 8 valence electrons on the outer shell so they can become stable.
For the following reaction, 5.29 grams of water are mixed with excess diphosphorus pentoxide. The reaction yields 13.3 grams of phosphoric acid . diphosphorus pentoxide(s) + water(l) phosphoric acid(aq). What is the theoretical yield of phosphoric acid?
Answer:
19.2 g
Explanation:
Step 1: Write the balanced equation
P₂O₅(s) + 3 H₂O(l) ⇒ 2 H₃PO₄
Step 2: Calculate the moles corresponding to 5.29 g of H₂O
The molar mass of H₂O is 18.02 g/mol.
5.29 g × 1 mol/18.02 g = 0.294 mol
Step 3: Calculate the theoretical yield of phosphoric acid, in moles
The molar ratio of H₂O to H₃PO₄ is 3:2. The theoretical yield of H₃PO₄ is 2/3 × 0.294 mol = 0.196 mol
Step 4: Calculate the mass corresponding to 0.196 moles of H₃PO₄
The molar mass of H₃PO₄ is 97.99 g/mol.
0.196 mol × 97.99 g/mol = 19.2 g
The correct geometry around oxygen in CH3OCH3 is
(a). linear. (b). bent. C). tetrahedral/(a). trigonal planar
Explanation:
the force of the lone pairs from the bottom would cancel out the force of the lone pairs from the top. Thus, the molecule will be linear.
Which of the following behaviors would best describe someone who is listening and paying attention? a) Leaning toward the speaker O b) Interrupting the speaker to share their opinion c) Avoiding eye contact d) Asking questions to make sure they understand what's being said
Answer:
a Leaning towards the speaker
5 points ) Which of the following is a benefit of using email to communicate at work ? a) You can express yourself in a limited number of characters b) You don't have to worry about using proper grammar. c) You always get a response right away. d ) You can reach a large audience with one communication .
Answer:
C
Explanation:
Urea, CH4N2O (s), is manufactured from NH3 (g) and CO2 (g). H2O (l) is another product of this reaction. An experiment is started with 2.6 grams of NH3 (g) added into a reaction vessel with CO2 (g).
Write the balanced equation for this reaction, being sure to include physical states. Based on the balanced equation above, calculate the following:
a. the theoretical yield of urea in grams that can be made from the NH3
b. the actual amount of urea made if the percent yield for this reaction is 34%.
Answer:
a. 4.41 g of Urea
b. 1.5 g of Urea
Explanation:
To start the problem, we define the reaction:
2NH₃ (g) + CO₂ (g) → CH₄N₂O (s) + H₂O(l)
We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:
2.6 g . 1mol / 17g = 0.153 moles of ammonia
Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea
0.153 moles ammonia may produce, the half of moles
0153 /2 = 0.076 moles of urea
To state the theoretical yield we convert moles to mass:
0.076 mol . 58 g/mol = 4.41 g
That's the 100 % yield reaction
If the percent yield, was 34%:
4.41 g . 0.34 = 1.50 g of urea were produced.
Formula is (Yield produced / Theoretical yield) . 100 → Percent yield
You are titrating 24.3 mL of 2.00 M HCl with 1.87 M NaOH. How much NaOH do you expect to have added when you reach the equivalence point?
26.0 mL
15.4 mL
13.4 mL
Answer:
26mL
Explanation:
NaOH+HCl= NaCl+H2O
nHCl=0.0243*2=0.0486
nNaOh=nHCl
VNaOH=0.0486/1.87=0.026l=26ml
Answer:
26.0 mL
Explanation:
Write a chemical equation for LiOH(aq) showing how it is an acid or a base according to the Arrhenius definition.
Answer:
LiOH(aq) → Li⁺(aq) + OH⁻(aq).
We know that water is purified before it is supplied to our houses. Then why do we have filters installed in our houses? What do they serve?
Answer:
Water filters remove elements that cause drinking water to have an unpleasant taste and smell, such as lead, chlorine and bacteria. Home water filtration system will improve the overall purity, taste and smell of your drinking water. It also lowers the pH level of the water that you drink.
How many grams of CO2 are formed if 44.7 g C5H12 is mixed with 108 g O2?
Explanation:
here's the answer to your question
PLEASE HELP I NEED THIS ASAP
Select all that apply.
The spectrum of Star S is compared to a reference hydrogen spectrum. What can be concluded about Star S?
Star S has radial motion.
Star S has transverse motion.
Star S is moving toward Earth.
Star S is moving away from Earth.
Answer:
I say Star S has radial motion
Explanation:
I'm not sure if it right but let me know if you have any other questions
Which of the following ions is the less likely to be formed?
A) Li+3
B) Na+
C) I-
D) Sr2+
Ε) Η+
Answer:
Li^3+
Explanation:
The electronic configuration of lithium is ; 1s2 2s1. This means that lithium has one electron in its outermost shell and two core electrons.
We know that it is difficult to remove these core electrons during ionization. Lithium belongs to group 1 hence Li^+ is formed more easily.
It is very difficult to form Li^3+ because it involves loss of core electrons which requires a lot of energy.
